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Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone in increasing when the height is 4 cm?

Manvendra Singh chahar , 10 Years ago
Grade Upto college level
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
Let height & radius of cone be at any instant ‘t’ to be ‘h’ & ‘r’.
Volume ‘V’:
V = \frac{1}{3}\pi r^{2}h
r = 6h
V = \frac{1}{3}\pi (6h)^{2}h
V = 12\pi (h)^{3}
\frac{\partial V}{\partial t} = 36\pi h^{2}.\frac{\partial h}{\partial t} = 12 \frac{cm^{3}}{sec}
h = 4cm
36\pi (4)^{2}.\frac{\partial h}{\partial t} = 12 \frac{cm^{3}}{sec}
\frac{\partial h}{\partial t} = \frac{1}{48\pi } \frac{cm}{sec}
Thanks & Regards
Jitender Singh
IIT Delhi
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